Answer:
Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2
 
        
             
        
        
        
Question:
A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q =  is located in the center of a spherical cavity of radius ,
 is located in the center of a spherical cavity of radius ,  m inside an insulating spherical charged solid.
  m inside an insulating spherical charged solid.  
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V  =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%2A%5Ctimes%2010%5E%7B-4%20%7D%5B8.57375%20-%202.81011%20%5D%5Ctimes%207.35%5Ctimes%2010%5E%7B-4%7D) 
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)


Electric field at  m due to shell
m due to shell
E1 =  

Electric field at   due to 'q' at center
 due to 'q' at center 
E2 =

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
= E2- E1
![=[  2.134  - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)

 
        
             
        
        
        
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²) 
of the force that you feel on the surface of the Earth.
That's <em>(1/18)</em> as much as on Earth.
The acceleration of gravity there would be about <em>0.545 m/s²</em>.  
This is about 12% less than the gravity on Pluto.